$ D = \left[\begin{array}{rr}2 & -1 \\ 0 & 1\end{array}\right]$ $ C = \left[\begin{array}{rr}1 & 3 \\ 3 & 1\end{array}\right]$ What is $ D C$ ?
Explanation: Because $ D$ has dimensions $(2\times2)$ and $ C$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ D C = \left[\begin{array}{rr}{2} & {-1} \\ {0} & {1}\end{array}\right] \left[\begin{array}{rr}{1} & \color{#DF0030}{3} \\ {3} & \color{#DF0030}{1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ C$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ C$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ C$ , and so on. Add the products together. $ \left[\begin{array}{rr}{2}\cdot{1}+{-1}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{1}+{-1}\cdot{3} & ? \\ {0}\cdot{1}+{1}\cdot{3} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{1}+{-1}\cdot{3} & {2}\cdot\color{#DF0030}{3}+{-1}\cdot\color{#DF0030}{1} \\ {0}\cdot{1}+{1}\cdot{3} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{2}\cdot{1}+{-1}\cdot{3} & {2}\cdot\color{#DF0030}{3}+{-1}\cdot\color{#DF0030}{1} \\ {0}\cdot{1}+{1}\cdot{3} & {0}\cdot\color{#DF0030}{3}+{1}\cdot\color{#DF0030}{1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-1 & 5 \\ 3 & 1\end{array}\right] $